Question

IV. Assume that IQ scores are normally distributed, with a mean of 100 and standard deviation of 15. What is the probability that a randomly selected person has an IQ score a greater than 120? b. less than 902 c. between 90 and 120? d. between 105 and 120?

Answer 1

a) P(X > 120)

= P((X - $\mu$)/$\sigma$ > (120 - $\mu$)/$\sigma$)

= P(Z > (120 - 100)/15)

= P(Z > 1.33)

= 1 - P(Z < 1.33)

= 1 - 0.9082

= 0.0918

b) P(X < 90)

= P((X - $\mu$)/$\sigma$ < (90 - $\mu$)/$\sigma$)

= P(Z < (90 - 100)/15)

= P(Z < -0.67)

= 0.2514

C) P(90 < X < 120)

= P((90 - $\mu$)/$\sigma$ < (X - $\mu$)/$\sigma$ < (120 - $\mu$)/$\sigma$)

= P((90 - 100)/15 < Z < (120 - 100)/15)

= P(-0.67 < Z < 1.33)

= P(Z < 1.33) - P(Z < -0.67)

= 0.9082 - 0.2514

= 0.6568

d) P(105 < X < 120)

= P((105 - $\mu$)/$\sigma$ < (X - $\mu$)/$\sigma$ < (120 - $\mu$)/$\sigma$)

= P((105 - 100)/15 < Z < (120 - 100)/15)

= P(0.33 < Z < 1.33)

= P(Z < 1.33) - P(Z < 0.33)

= 0.9082 - 0.6293

= 0.2789